Given that angle BAC = 37^ and angle ACD = 46^.

Angle x is $\angle ABC$.

Since AC is the diameter, the angle subtended by the diameter at any point on the circumference is $90^\circ$.
Therefore, $\angle ABC = 90^\circ$.

The value of angle x is $\boxed{90^\circ}$.

Angle y is $\angle CAD$.

Since AC is the diameter, the angle subtended by the diameter at point D on the circumference is $90^\circ$.
Therefore, $\angle ADC = 90^\circ$.

In $\triangle ADC$, the sum of angles is $180^\circ$.

The value of angle y is $\boxed{44^\circ}$.

Angle DCB is the sum of $\angle DCA$ (which is $\angle ACD$) and $\angle ACB$.

Consider $\triangle ABC$. We know $\angle ABC = 90^\circ$ (from part a) and $\angle BAC = 37^\circ$ (given).
The sum of angles in $\triangle ABC$ is $180^\circ$.

Angle DCB is $\boxed{99^\circ}$.